Chapter 2.8, Problem 70E

Solution

To solve: The problem without using a calculator.

The solution set is equal to $\left[-1,1\right]$ .

Given information:

Consider the given inequality.

  ${\left(x^2-1\right)}^2\le 0$

Calculation:

To set numerator and the denominator equal to zero. These are the values that make the quotient zero or undefined and solve for $x$ :

  $\begin{array}{c}{\left(x^2-1\right)}^2=0\\ x^2-1=0\\ x^2=1\\ x=\pm 1\end{array}$

So, the boundary points are $-1,1$ .

Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

The boundary points divide the number line into three test intervals. Including the boundary points (because of the given less than sign), the intervals are below:

  $\left(-\infty ,-1\right],\left[-1,1\right],\left[1,\infty \right)$

Now, take one representative number within each test interval and substitute that number into the inequality below:

  $\begin{array}{l}\left(-\infty ,-1\right],\left[-1,1\right],\left[1,\infty \right)\\ {\left({\left(-2\right)}^2-1\right)}^2\le 0{\left(1^2-1\right)}^2\le 0{\left(2^2-1\right)}^2\le 0\\ 9\le 0\text{False }0\le 0\text{True}9\le 0\text{False}\end{array}$

Therefore, the intervals $\left[-1,1\right]$ belongs to the solution set and the intervals $\left(-\infty ,-1\right]$ and $\left[1,\infty \right)$ does not belong to the solution set.

Hence, the solution set is $\left[-1,1\right]$ .